"If E = 7.50V and r=0.45Ω, find the minimum value of the voltmeter resistance RV for which the voltmeter reading is within 1.0%

Question

3 months ago

10

of the emf of the battery. Express your answer numerically (in ohms) to at least three significant digits.

1 answer:

ValentinkaMS [3.4K] · Answer 1 · 2026-03-06T00:44:16+03:00

Answer:

The minimum resistance value is $R_V =44.552\ \Omega$

Explanation:

According to the question, we have:

The voltage given as $E = 7.5V$

The internal resistance as $r = 0.45$

The goal here is to find the minimum resistance for the voltmeter such that its reading is within 1.0% of the battery's emf.

This means we require voltmeter resistance such that:

V = (100% - 1%) of E

Where E is the battery's e.m.f. and V is the voltmeter reading.

So, V = 99% of E = 0.99 E = 7.425.

In general, we have:

E = V + ir

where ir denotes the internal voltage drop across the voltmeter, and V is the reading from the voltmeter.

By rearranging, we get:

$i = \frac{(E-V)}{r}$

$=\frac{7.50-7.425}{0.45}$

$= 0.1667 A$

Since the current remains constant throughout the circuit:

$V = iR_V$

where $R_V$ is the voltmeter resistance value.

Hence, $R_V = \frac{V}{i} = \frac{7.425}{0.1667}$

$=44.552\ \Omega$

1/7 kg